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In my quest on how to select the relevant TVS diode, I came across this document.

Page 3 paragraph 2.1, it is stated :

Vrwm

The reverse working maximum voltage (VRWM) is defined as the voltage that can be applied to a TVS diode with an assurance that the diode will not, over process or temperature, conduct significant current. The definition of ‘significant current’ depends on the TVS manufacturer, but generally it is < 100 nA. The VRWM specification enables a designer to select a device that will see minimal leakage across all operating conditions. In a design, VRWM should be selected to ensure that it is above the expected maximum operating voltage. If the applied voltage rises above VRWM, there is a chance to see diode leakage increase significantly. For example, if the protected line operates at 5 V nominal with a maximum variance up to 7 V, ensure that the VRWM is 7 V or greater.

I don't understand why the last sentence is exact:

If nominal operation of device is 5 V, with maximum 7 V. I understand that if the voltage raises above 7 V, the device will die. So, If I choose a TVS with a Vrmw of 8 V (as noted in the last sentence "ensure that the VRWM is 7 V or greater") and the voltage goes to 7.5 V, my interpretation is that the diode will not conduct significantly (because voltage is below Vrmw) and my device will die.

What's wrong with my understanding?

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  • \$\begingroup\$ You don't (or shouldn't) rate devices right to the edge of their breaking point. You don't rate a bridge at 1000 tons if it will snap at 1001 tons. \$\endgroup\$
    – DKNguyen
    Oct 6, 2021 at 15:18
  • \$\begingroup\$ So you mean that when a manufacturer says "my component is rated 30 V", in fact, it's more? right ? \$\endgroup\$
    – Julien
    Oct 6, 2021 at 16:11
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    \$\begingroup\$ When they say it's rated for 30V, they really are rating it for 30V, by definition. A rating is not the same as the absolute limit. They won't guarantee anything beyond rated and you shouldn't leave that region on purpose. \$\endgroup\$
    – DKNguyen
    Oct 6, 2021 at 18:47

1 Answer 1

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The key word is "operating conditions". In this example, the device is 5V nominal, but it can withstand 7V without immediate damage, and it might occasionally see a 7V supply. Then the TVS should be selected so that it doesn't conduct at 7V (it would be a bad day for everyone involved if it did).

The TVS Vrwm should be chosen somewhere in the zone between the highest "expected" supply voltage and the voltage that causes damage. If there isn't any margin between those values, then you have a problem that a TVS won't fix.

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  • \$\begingroup\$ ok. For example st.com/resource/en/datasheet/stm32l072cz.pdf . page 64, VDD max 3.6V. Page 62 absolute maximum rating for VDD is 4.0V. Based on your answer, I understand that Vrwm should be higher than 4.0V. How do I estimate it? a rule of thumb of few volts more? Can I find a relevant information in datasheet? Paragraph 6.3.11 deals with ESD. I can see 500V. May be that's only for very fast transcient. \$\endgroup\$
    – Julien
    Oct 6, 2021 at 16:07
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    \$\begingroup\$ @Julien no. I'm not saying that, unless you plan on using this with an input that has 4.0V within its normal range of variation (which would be silly). Realistically you would choose a 3.3V nominal voltage, a 3.6V Vrwm, and the lowest Vbr you can reasonably get. Assuming a TVS is even appropriate here. \$\endgroup\$
    – hobbs
    Oct 6, 2021 at 16:27
  • \$\begingroup\$ I think I get what you mean. So, in the above example, Ideally both Vrwm and Vbr should be below 4.0V? Right? \$\endgroup\$
    – Julien
    Oct 7, 2021 at 10:03

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